Leetcode 470. 用 Rand7() 实现 Rand10()（拒绝采样）

2021年03月19日 周五 天气晴 【不悲叹过去，不荒废现在，不惧怕未来】

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1. 问题简介

470. 用 Rand7() 实现 Rand10()

2. 题解（拒绝采样）

(randX() - 1)*Y + randY() 可以等概率的生成[1, X * Y]范围的随机数。

具体可参考：从最基础的讲起如何做到均匀的生成随机数

2.1 方法一

class Solution {
    
    
public:
    int rand10() {
    
    
        while(true){
    
    
            int a = rand7();
            int b = rand7();
            int num = (a-1)*7 + b; // rand 49
            if(num <= 40) return num % 10 + 1; // 拒绝采样
            
            a = num - 40; // rand 9
            b = rand7();
            num = (a-1)*7 + b; // rand 63
            if(num <= 60) return num % 10 + 1;
            
            a = num - 60; // rand 3
            b = rand7();
            num = (a-1)*7 + b; // rand 21
            if(num <= 20) return num % 10 + 1;
        }
    }
};

期望计算：

A = 2 + 9/49 * (1 + 3/63 * 1) = 2.192419825072886
p = 9/49 * 3/63 * 1/21 = 0.00041649312786339027
EX = A + A * p + A * p^2 + ... = A / (1 - p) = 2.1933333333333334

2.2 方法二

class Solution {
    
    
public:
    int rand10() {
    
    
        while(true){
    
    
            int a = rand7(), b = rand7();
            while(a == 7) a = rand7();   //让a只能为1到6，保证奇数偶数个数相同
            while(b > 5) b = rand7();    //让b只能为1到5
            return (a & 1 ? 0 : 5) + b;  //相当于将10分解为两个部分，1/2 * 1/5 = 1/10
        }
    }
};

期望计算：

E1 = 2 (int a = rand7(), b = rand7();)
E2 = 1/7 * 1 + 1/7 * 1/7 * 1 + 1/7 * 1/7 * 1/7 * 1... =  1/6 (while(a == 7) a = rand7();)
E2 = 2/7 * 1 + 2/7 * 2/7 * 1 + 2/7 * 2/7 * 2/7 * 1... =  2/5 (while(b > 5) b = rand7();)
E = E1 + E2 + E3 = 2 + 1/6 + 2/5 = 2.566666666666667

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参考文献

https://leetcode-cn.com/problems/implement-rand10-using-rand7/solution/yong-rand7-shi-xian-rand10-by-leetcode/

https://leetcode-cn.com/problems/implement-rand10-using-rand7/solution/cong-zui-ji-chu-de-jiang-qi-ru-he-zuo-dao-jun-yun-/